Question

The ratio of the number of moles of $AgN{O}_3$, $Pb(N{O}_3{)}_2$ and $Fe(N{O}_3{)}_3$ required for coagulation of a definite amount of a colloidal sol of silver iodide prepared by mixing $AgN{O}_3$ with excess of KI will be :

• A. 1 : 2 : 3
• B. 3 : 2 :1
• C. 6 : 3 : 2
• D. 2 : 3 : 6

Answer 1

The correct option is C 6 : 3 : 2

With axcess of $KI$, colloidal particles will be $\left[AgL\right]I^{-}$
$\underset{1mol}{\overset{\left[AgL\right]I^{-}}{\to }}+\underset{1mol}{\overset{AgN{O}_3}{\to }}\to AgI\downarrow +AgI\downarrow +N{O}_3^{-}$
$\underset{2mol}{\overset{2\left[AgL\right]I^{-}}{\to }}+\underset{1mol}{\overset{Pb(N{O}_3{)}_2}{\to }}\to 2AgI\downarrow +Pb{I}_2\downarrow +2N{O}_3^{-}$
$\underset{\begin{array}{c}3mole\\ 1mole\end{array}}{3\left[AgI\right]I^{-}}+\underset{\begin{array}{c}1mole\\ mole\end{array}}{Fe(N{O}_3{)}_3}\to 3AgI\downarrow +Fe{I}_2\downarrow +3N{O}_3^{-}$
$\therefore$ Molar ratio required for coagulation of same amount of $\left[AgI\right]I^{-}$ is $=1:\frac{1}{2}:\frac{1}{3}=6:3:2$